A light spring obeys Hooke's law. The spring's unstretched length is 34.0 cm. One end of the spring is attached to the top of a doorframe and a weight with mass 7.00 kg is hung from the other end. The final length of the spring is 44.5 cm. a) Find its spring constant.
1. kN/m
(b) The load and the spring are taken down. Two people pull in opposite directions on the ends of the spring, each with a force of 150 N. Find the length of the spring in this situation.

Explanation:

Hooke's law is represented by thee formula

F = ke where F is force in N and K is the spring constant.

Initial length of the spring = 34cm = 0.34 m

mass of 7.00kg hung

weight = mg = 7 × 9.8 = 68.6 N

Final length of the spring = 44.5 cm = 0.445 m

extension = final length - initial length = 0.445 m - 0.34 m = 0.105 m

a) F = Ke

K = F / e = 68.6 N /0.105 m = 653.33 N/m = 0.653 kN/m

b) F = 150 N

k = 653.33 N/m

F = ke

150 N / 653.33 N/m = e

e = 0.23 m

new length = 0.34 + 0.23= 0.57 m = 57 cm

i think if it were me i whould divide 300 dide 15

step-by-step explanation: